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Yahoo!

  • 3 years ago

(cos A + cos b)^2 + (sin A + sin B)^2=?

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  1. Yahoo!
    • 3 years ago
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    @apoorvk @satellite73 @A.Avinash_Goutham @ajprincess @AndrewNJ

  2. apoorvk
    • 3 years ago
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    okay, expand the two whole-squares, it's simple enough...

  3. Yahoo!
    • 3 years ago
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    the options are 4cos^2(a-b/2) 4 sin a/2 cos b/2

  4. Yahoo!
    • 3 years ago
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    |dw:1339426035616:dw|

  5. Yahoo!
    • 3 years ago
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    @apoorvk after expanding i got this then next..............

  6. apoorvk
    • 3 years ago
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    great, now.. cosAcosB + sinAsinB = cos(A-B) Use this^

  7. patmyer1
    • 3 years ago
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    ugh, trig identities are a drag. Even after deriving them myself I simply CANNOT seem to memorize them

  8. AndrewNJ
    • 3 years ago
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    spread it so we have (cosa)^2+(cosb)^2+2(cosa)(cosb)+(sina)^2+(sinb)^2+2(sina)(sinb) =2+2cos(a-b)

  9. AndrewNJ
    • 3 years ago
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    (cosa)^2+(sina)^2=1 (cosb)^2+(sinb)^2=1 (cosa)(cosb)+(sina)(sinb)=cos(a-b)

  10. Yahoo!
    • 3 years ago
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    ok @apoorvk so it is not over yet!!@

  11. shivam_bhalla
    • 3 years ago
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    Continuing from where @AndrewNJ left, \[2(1+\cos(a-b))\] now we know that \[1 + \cos 2a = 2 \cos^2 a\] Now try :D

  12. Yahoo!
    • 3 years ago
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    @shivam_bhalla did nt understand!

  13. shivam_bhalla
    • 3 years ago
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    = \[2(1+\cos 2\frac{(a-b)}{2})\] Now we have a identity, \[1 + \cos 2x = 2 \cos^2 x\] in your case, \[x = \frac{a-b}{2}\] Now try

  14. Yahoo!
    • 3 years ago
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    i got it thxx @shivam_bhalla

  15. shivam_bhalla
    • 3 years ago
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    Your welcome :)

  16. Yahoo!
    • 3 years ago
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    @shivam_bhalla plzz help http://openstudy.com/study?signup#/updates/4fd60c0be4b04bec7f170b7d

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